Problem: Simplify, then add the following rational expressions. You may keep the denominator in its factored form. You may also assume that no denominators equal zero. $\dfrac{6z^2-12z}{4z^2-16z+16}+\dfrac{3z}{z^2-z-2}=$
Let's start by factoring the first rational expression: $\dfrac{6z^2-12z}{4z^2-16z+16}=\dfrac{6z(z-2)}{4(z-2)(z-2)}=\dfrac{3z}{2(z-2)}$ Now let's factor the second rational expression: $\dfrac{3z}{z^2-z-2}=\dfrac{3z}{(z+1)(z-2)}$ Now the addition problem looks as follows: $\dfrac{3z}{{2}{(z-2)}}+\dfrac{3z}{{(z-2)}{(z+1)}}$ We can add two rational expressions whose denominators are equal by adding the numerators and keeping the denominator the same. When the denominators are not the same, we must manipulate them so that they become the same. More specifically, we should find the least common multiple of the two denominators. [What's that?] We can see that: Both denominators share the factor ${(z-2)}$. Only the first denominator has the factor ${2}$. Only the second denominator has the factor ${z+1}$. Therefore, the least common multiple is the product of all the above factors. Let's manipulate the first expression to have that denominator: $\begin{aligned}&\phantom{=}\dfrac{3z}{{2}{(z-2)}}+\dfrac{3z}{{(z-2)}{(z+1)}}\\\\\\\\ &=\dfrac{3z{(z+1)}}{{2}{(z-2)}{(z+1)}}+\dfrac{3z{(2)}}{{(z-2)}{(z+1)}{(2)}}\\\\\\\\ &=\dfrac{3z^2+3z}{2(z+1)(z-2)}+\dfrac{6z}{2(z+1)(z-2)}\end{aligned}$ Now that both denominators are the same, let's add! $\phantom{=}\dfrac{3z^2+3z}{2(z+1)(z-2)}+\dfrac{6z}{2(z+1)(z-2)}$ $\begin{aligned} &=\dfrac{(3z^2+3z)+(6z)}{2(z+1)(z-2)}&\text{Add numerators}\\\\\\\\ &=\dfrac{3z^2+3z+6z}{2(z+1)(z-2)}&\text{Distribute}\\ &=\dfrac{3z^2+9z}{2(z+1)(z-2)}&\text{Combine like terms}\\\end{aligned}$ In conclusion, $\dfrac{6z^2-12z}{4z^2-16z+16}+\dfrac{3z}{z^2-z-2}=\dfrac{3z^2+9z}{2(z+1)(z-2)}$